Consider a fixed charge Q1. Let us move a second charge say test charge Qt around the Q1. We observe that the test charge Qt experiences force everywhere around Q1. It is experiencing a force field. Force on it will be given by coulombs law as:

But still question arises what is Electric Field?? Before defining electric field let us modify above equation a bit.
Let us write the above equation as force/charge equation:

This quantity that is force/charge describes a vector quantity. This vector quantity is called Electric Field Intensity.So Electric field intesity is defined as the vector force on an unit positive test charge by a fixed charge Q1. Or it can also be defined as force per unit charge.
So to sum it up we can Electric field intensity as:

So it becomes.

Expressing Electric field in cartesian Coordinate:
![E(r) = \frac{Q}{4\pi \epsilon _{0}\left | r - r^{'} \right |^{2}}\frac{r - r^{'}}{\left | r - r^{'} \right |} = \frac{Q[(x - x^{'})i + (y - y^{'})j + (z - z^{'})k]}{4\pi \epsilon _{0}[(x - x^{'})^{2} + (y - y^{'})^{2} + (z - z^{'})^{2}]^{\frac{3}{2}}}](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u9ihTmCxF3yfOylBQmvqXrh5bscmWIxa5FuaJyt-YbaHXZHJnUsEWEDu47YkCxD3LHTCod4PWkN5V8zPiQ5DSeuAXQt_vS4kFiWjqy1QBsypO9jn8e2O_Bz0sbJULdP_l9xoGCGwEWr3E2GNJH44U0NeqXb1R7x7LwxVhk6_PmiLk7Fl2B19NuYXwpBA8x-gn5yRBkpgOQsnJbT9OlOfweDWU2zEcyqy9tBefBllcb7OUwu2UPCPY8mFHwNxSx_DICbbFZYjvf3azFOJ29H0K8CJjy-xQJIft5u8nnMaF8ILuOJyaUVV4Y3xghcnTp_PyCrMAC-9BYeFXfknnX8Z5ZApZKAohnl5wOAdjTt6C_gF_0eKGssn2RQktnMS-jU04oDUyo7-ViiDAH82zzBpLIYw_vG54gJfXI4kv-dd2peAhFZxKLar62_VSFP6yXwQfnAZG8CKAK2BBHJsdnvuEUVhZ0jHnoVL3JeF8NnANm5PLHdy3FWz7h_e6v6y7xGc6THLenwQFtfThbBR6UApza_okluibeKH2mOHLmotFurMkYDuaG3GvGS2spcNPioSKhtX_64P4mHj1uN5PcDSp_h2L9EelY8PX1UMn4zsqjMFc-9SqlISo6jt2Szdz5ozXJMGcQbrSa58Y=s0-d)
Expressing Electric Field in Spderical Co-ordinates:
First for considering spherical co ordinate we must see that fixed charge has spherical symmetry at its location like at origin(0,0,0) for that we can either use spherical co-ordinate directly in equation as:

Where Er and r spherical co-ordinates values or we can convert them in cartesian coordinate which will give us the equation as below:
We will have r = R = xi + yj +zk ar = aR = (xi + yj + zk)/√(x^2+y^2+z^2 )

If more charges are added in the surrounding of Q1 then electric field will be sum of individual acting alone. As shown in equation given below:

Before ending the topic let us do a small quick example.
Example: See the figure below. We need to find out Electric field at P(1,1,1) caused by 4 identical 3nC charges located at P1(1,1,0), P2(-1,1,0), P3(-1,-1,0) and P4(1,-1,0).
[caption id="" align="alignnone" width="615" caption="Example of Electric Field"][/caption]
Answer: As r = i + j + k and r1 = i+j therefore r – r1 = k. Therefore |r – r1| = 1.
Similaraly r – r2 = 2i + k. therefore |r – r2| = √5
Similaraly r – r3 = 2i + 2j + k. therefore |r – r3| = 3
Similaraly r – r4 = 2j + k. therefore |r – r4| = √5
And as

![E = 26.96[\frac{k*1}{1*1^{2}}+\frac{2i + k}{\sqrt{5}}\frac{1}{\sqrt{5}^{2}} + \frac{2i + 2j + k}{3}\frac{1}{3^{2}}+ \frac{2j + k}{\sqrt{5}}\frac{1}{\sqrt{5}^{2}}]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tTiMIh90zW_gQ8J7Jm9UYSG6B-LvHiu6V6YQ4NXHVjj1DjF-K0m_3ZroDdIphxaxkyFS1mhWsrfe6HcpCWX6JblG58WiL8-9DatlLT9WX4Lnj6e12AFLtxZ4pwAYFVTfosRluJFDKqtCry3DThelGiRobjEjecwRIO4Fg929N_FgOE5l8N8x82lkMvu0JOQRTJK6QJdulD2_xMnf6mnWKgQzIUcnzPF-FGjLS3PZHtW3IGzL6Io5WUt_-z6WKgNMnfB4_1Tj6lpDteFk0HjDj_HssFTWV2CREvEwE7pYh6QXMkErLgBfEutKC0MzsX9WWZxpnUZUhJnSUxkD8DW_i58gnWC59ugVWGpCjLWMpJONLECkNuhZJkCVCrdoycc38lgoaxOS8i_bs5MYZC9HtOkKYohmTKB6Rjn7memD1Wx2CUHDgQX5mJvY1t9iSyAVIuKIGObhDW3Os3AToQMyHHZAtmCw=s0-d)
Therefore simplifying we get E =
E = 6.82i + 6.83j + 32.8k V/m ans.
But still question arises what is Electric Field?? Before defining electric field let us modify above equation a bit.
Let us write the above equation as force/charge equation:
This quantity that is force/charge describes a vector quantity. This vector quantity is called Electric Field Intensity.So Electric field intesity is defined as the vector force on an unit positive test charge by a fixed charge Q1. Or it can also be defined as force per unit charge.
So to sum it up we can Electric field intensity as:
So it becomes.
Expressing Electric field in cartesian Coordinate:
Expressing Electric Field in Spderical Co-ordinates:
First for considering spherical co ordinate we must see that fixed charge has spherical symmetry at its location like at origin(0,0,0) for that we can either use spherical co-ordinate directly in equation as:
Where Er and r spherical co-ordinates values or we can convert them in cartesian coordinate which will give us the equation as below:
We will have r = R = xi + yj +zk ar = aR = (xi + yj + zk)/√(x^2+y^2+z^2 )
If more charges are added in the surrounding of Q1 then electric field will be sum of individual acting alone. As shown in equation given below:
Before ending the topic let us do a small quick example.
Example: See the figure below. We need to find out Electric field at P(1,1,1) caused by 4 identical 3nC charges located at P1(1,1,0), P2(-1,1,0), P3(-1,-1,0) and P4(1,-1,0).
Example of Electric field |
Answer: As r = i + j + k and r1 = i+j therefore r – r1 = k. Therefore |r – r1| = 1.
Similaraly r – r2 = 2i + k. therefore |r – r2| = √5
Similaraly r – r3 = 2i + 2j + k. therefore |r – r3| = 3
Similaraly r – r4 = 2j + k. therefore |r – r4| = √5
And as
Therefore simplifying we get E =
E = 6.82i + 6.83j + 32.8k V/m ans.
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